Two charges are placed on the x axis. One of the charges (q1 = +7.91C) is at x1 = +3.00 cm and the other (q2 = -27.3C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

see above question.

To find the net electric field at a given point, you need to calculate the electric fields produced by each individual charge and then add them vectorially. The formula to calculate the electric field at a point due to a point charge is given by:

Electric Field (E) = (k * q) / r^2

where k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2), q is the charge, and r is the distance between the charge and the point where you want to find the electric field.

Let's calculate the electric field at each point:

(a) x = 0 cm:
We have two charges: q1 = +7.91C and q2 = -27.3C.
The distance between q1 and x = 0 cm is r1 = 3.00 cm = 0.03 m.
The distance between q2 and x = 0 cm is r2 = 9.00 cm = 0.09 m.

The electric field produced by each charge at x = 0 cm is given by:
E1 = (k * q1) / r1^2
E2 = (k * q2) / r2^2

To find the net electric field, we need to add these two electric fields, taking into account their signs.

(b) x = +6.00 cm:
We use the same formula to calculate the electric fields produced by each charge at x = 6.00 cm. The distances will be different, but the calculations are similar to part (a).

Now, let's plug in the values and calculate the electric fields at the given points:

(a) x = 0 cm:
E1 = (9.0 x 10^9 N m^2/C^2 * 7.91 C) / (0.03 m)^2
E2 = (9.0 x 10^9 N m^2/C^2 * -27.3 C) / (0.09 m)^2

To find the net electric field, we add these two electric fields:
Net Electric Field at x = 0 cm = E1 + E2

(b) x = +6.00 cm:
You can follow the same procedure as in part (a) to calculate the electric field at x = 6.00 cm using the formula and values provided.

By following these steps, you will be able to calculate the net electric field (magnitude and direction) at the specified points.